Answer:
0
Step-by-step explanation:
-9x -3(2) = 6
-9x + 6 = 6
-9x + 6 -6 = 6-6
-9x/ -9= 0/-9
x=0
<span>Here let the quadratic equation be ax^2 + bx + c
We know that a=5 from the question.
Since the roots are 6 and 2, the quadratic equation would take the form of a product like (a1x-b1)(a2x-b2).
However, let's assume that a2=1 and b2=6,
Since a=5, a1=5, then 5x-b1=5(x-2). Solving this shows that b1=10
So, the equation is (5x-10)(x-6)</span>
Answer:
(m) =
ΔY
ΔX
= 0
Step-by-step explanation:
Check the picture below.
so we know the radius of the semicircle is 2 and the rectangle below it is really a 4x4 square, so let's just get their separate areas and add them up.
![\stackrel{\textit{area of the semicircle}}{\cfrac{1}{2}\pi r^2}\implies \cfrac{1}{2}(\stackrel{\pi }{3.14})(2)^2\implies 3.14\cdot 2\implies 6.28 \\\\\\ \stackrel{\textit{area of the square}}{(4)(4)}\implies 16 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{sum of both areas}}{16+6.28=22.28}~\hfill](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20semicircle%7D%7D%7B%5Ccfrac%7B1%7D%7B2%7D%5Cpi%20r%5E2%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B2%7D%28%5Cstackrel%7B%5Cpi%20%7D%7B3.14%7D%29%282%29%5E2%5Cimplies%203.14%5Ccdot%202%5Cimplies%206.28%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20square%7D%7D%7B%284%29%284%29%7D%5Cimplies%2016%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20both%20areas%7D%7D%7B16%2B6.28%3D22.28%7D~%5Chfill)